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SVEN [57.7K]
1 year ago
11

A hammer is used to hit a nail into a block of wood. The hammer hits the nail with a speed of 8.0 m/s and then stops. The hammer

is in contact with the nail for 0.0015 s.,hammer has mass 0.15 kg.Calculate the average force between the hammer and the nail.
Physics
1 answer:
77julia77 [94]1 year ago
6 0

800 Newtons

Explanation

The average force is the force exerted by a body moving at a defined rate of speed (velocity) for a defined period of time.

the average force is given by:

F=ma

and

a=\frac{\Delta v}{\Delta t}

\begin{gathered} F_{average}=m\frac{\Delta v}{\Delta t} \\ where\text{ m is the mass of the objectt} \\ \Delta v\text{ is the change in velocity} \\ \Delta t=\text{ time} \end{gathered}

Step 1

a) Let

\begin{gathered} m=0.0015\text{ kg} \\ \Delta v=0.0015s \\ \Delta v=8\text{ }\frac{m}{s} \end{gathered}

now, replace

\begin{gathered} F=0.15\text{ kg}\frac{0-8\frac{m}{s}}{0.0015\text{ s}} \\ F=-800\text{ N} \end{gathered}

the negative sign indicates the force is in the opposite way ( the force is exerted by the nail to the hammer), so the force is opposite to the direction of the movement

so, the answer is

800 Newtons

I hope this helps you

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When water waves meet, they can combine to form new waves. In constructive waves, a ________ amplitude wave is formed. In destru
Sever21 [200]

Answer:

In constructive waves, a <u><em>greater</em></u> amplitude wave is formed. In destructive waves, a wave with a <u><em>smaller</em></u> amplitude is formed. (option A)

Explanation:

Interference is called the superposition or sum of two or more waves. Depending mainly on the wavelengths, amplitudes and the relative distance between them, there are two types of interference: constructive or destructive.

Constructive interference occurs when there are two waves of identical or similar frequency (both have motions equal to an even number of similar wavelengths) and overlap the peak of one with the peak of the other. These effects add together and make a wave of greater amplitude. All of this is possible because the waves were in the same phase in the beginning (in the same position).

Destructive interference occurs in the opposite case to constructive. When the crest of one wave overlaps the valley of the other, they cancel out since they are in different phases when they overlap (they were in different positions). That is, as in the case of constructive waves they were added, in the case of destructive waves they cancel out (subtract).

So, <u><em>In constructive waves, a greater amplitude wave is formed. In destructive waves, a wave with a smaller amplitude is formed. </em></u>

3 0
3 years ago
A typical land snail's speed is 12.2 meters per hour. How many miles will the snail travel in one day(24hrs)?
algol13
The snail will go <span>0.18193752 miles </span>
5 0
3 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

7 0
3 years ago
You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend
Bogdan [553]

Answer:

a.\tau=200J b.\alpha=0.44 \frac{rad}{s^2} c. \alpha=0.33\frac{rad}{s^2} d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by \tau=rF\sin \theta, being r the radius, F the force aplied and \theta the angle between the vector force and the vector radius. Since \theta=90^{\circ}, \, \sin\theta=1 and so \tau=rF=2m100N=200Nm=200J.

b. Since the relation \tau=I\alpha hols, being I the moment of inertia, the angular acceleration can be calculated by \alpha=\frac{\tau}{I}. Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is I=\frac{1}{2}Mr^2, being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by I_p=m_pr_p^{2}, being m_p the mass of the person and r_p^{2} the distance from the person to the center. Given all of this, we have

\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}.

c. Similar equation to b, but changing r_p=2m, so

alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}.

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0
3 years ago
a man hits a golf ball (0.2kg) which accelerates at a rate of 20 m/s what amount of force acted on the ball
MAVERICK [17]

The ball only accelerates during the brief time that the club is in contact
with it. After it leaves the club face, it takes off at a constant speed.

If it accelerates at 20 m/s² during the hit, then

   Force = (mass) x (acceleration) = (0.2kg) x (20 m/s²) = <em>4 newtons</em> .


8 0
3 years ago
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